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Question : 24 of 160
Marks:
+1,
-0
Solution:
cos‌6‌x+cos‌4‌x+cos‌2‌x=−1 in
[0,π] ⇒‌‌(1+cos‌6‌x)+(cos‌4‌x+cos‌2‌x)=0 ⇒‌‌{1+cos‌2(3x)}+(2‌cos‌3‌x⋅cos‌x)=0 ⇒‌‌(1+2cos23x−1)+2‌cos‌x‌cos‌3‌x=0 ⇒‌‌2‌cos‌3‌x(cos‌3‌x+cos‌x)=0 ‌‌{∵cos‌C+cos‌D=2‌cos‌⋅cos‌} ⇒‌‌2‌cos‌3‌x‌2⋅cos‌2‌x⋅cos‌x=0 ⇒‌‌cos‌x⋅cos‌2‌x‌cos‌3‌x=0 ∴x=30∘,45∘,90∘,135∘,180∘ in
[0,Ï€]
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